# QFT: Computing S-Operator to 1st Order in Coupling Constant lambda

• WarnK
In summary, Wick's theorem tells you to sum over all the possible contractions so you can contract the first field with the second and the third with the fourth or the first with the third and the second with the third. each phi is a linear combination of creation and annihilation terms, so it's not too hard to believe that you can get some non-zero piece out of <phi^4>.
WarnK

## Homework Statement

Compute the S-operator to first order in the coupling constant lambda.

## Homework Equations

The given Lagrangian density is
$$L = : \frac{1}{2} (\partial_{\mu} \phi)^2 - \frac{1}{2}m^2\phi^2 + \frac{1}{2}\frac{\lambda}{4!}\phi^4 :$$
where phi is a scalar field.

## The Attempt at a Solution

S = 1+iT
and I want to calculate iT to first order, which I guess is
$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
using Wick's theorem, how is this anything except zero? Or I'm I missing something?

WarnK said:

## Homework Statement

Compute the S-operator to first order in the coupling constant lambda.

## Homework Equations

The given Lagrangian density is
$$L = : \frac{1}{2} (\partial_{\mu} \phi)^2 - \frac{1}{2}m^2\phi^2 + \frac{1}{2}\frac{\lambda}{4!}\phi^4 :$$
where phi is a scalar field.

## The Attempt at a Solution

S = 1+iT
and I want to calculate iT to first order, which I guess is
$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
using Wick's theorem, how is this anything except zero? Or I'm I missing something?

Why do you think this should give zero?Wick's theorem tells you to sum over all the possible contractions so you can contract the first field with the second and the third with the fourth or the first with the third and the second with the fourth or the first with the fourth and the second with the third.

WarnK said:
how is this anything except zero?
each phi is a linear combination of creation and annihilation terms, so it's not too hard to believe that you can get some non-zero piece out of <phi^4>. To say something quantitative you want to write the time ordered product in terms of normal ordered products and contractions... which is done using wick's theorem.

$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
is only one kind of diagram; a disconnected one with no external legs. And that doesn't contribute to any scattering?

WarnK said:
$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
is only one kind of diagram; a disconnected one with no external legs. And that doesn't contribute to any scattering?

but that doesn't mean it is equal to zero. The disconnected diagrams in a scattering amplitude factor and cancel with the "denominator" of, say <T[phi_x phi_y phi_z phi_w S(inf)]>/<S(inf)>, but they aren't zero, they just don't "contribute" to the amplitude.

if you want to calculate, say, the free energy then you have to evaluate diagram w no external legs explicitly.

WarnK said:
$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
is only one kind of diagram; a disconnected one with no external legs. And that doesn't contribute to any scattering?

You are right. Is this really the calculation you have to do?

Normally, if you have to calculate the scattering of, say, two particles to two particles, you will need to evaluate not th eexpression you wrote above but rather

$$<0|T\big( -i ~ \phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4) \int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$

i.e. there are fields connected to the external spacetime points. Are you sure you don't have those fields as well? If not, then you get only disconnected diagrams as you said. And it makes sense that there is no scattering from your expression since there is no extrenal field to connect to!

## 1. What is the purpose of computing the S-operator to 1st order in the coupling constant lambda in QFT?

The S-operator is a fundamental concept in quantum field theory (QFT) that describes the evolution of a quantum state over time. By computing the S-operator to 1st order in the coupling constant lambda, we can understand the first-order effects of interactions between particles in a quantum system.

## 2. How is the S-operator related to Feynman diagrams in QFT?

Feynman diagrams are graphical representations of calculations involving the S-operator in QFT. Each line in a Feynman diagram corresponds to a term in the S-operator, and the vertices represent interactions between particles. Computing the S-operator to 1st order in lambda allows us to understand the lowest order contributions to these Feynman diagrams.

## 3. What is the mathematical expression for the S-operator to 1st order in lambda?

The mathematical expression for the S-operator to 1st order in lambda is given by the perturbative expansion, which is a sum of terms involving the coupling constant lambda and the corresponding Feynman diagrams. This expression can be solved using various techniques, such as perturbation theory or the path integral formulation.

## 4. How does the value of lambda affect the calculation of the S-operator?

The value of lambda determines the strength of the interactions between particles in a quantum system. As we compute the S-operator to higher orders in lambda, we take into account more and more interactions between particles, which can significantly impact the overall behavior of the system. The value of lambda can also affect the convergence of the perturbative expansion.

## 5. How does computing the S-operator to 1st order in lambda contribute to our understanding of QFT?

Computing the S-operator to 1st order in lambda is a crucial step in understanding the behavior of quantum systems in QFT. It allows us to make predictions about the interactions between particles and can help us verify the validity of theoretical models. By studying the S-operator at different orders, we can gain a deeper understanding of the fundamental principles of quantum mechanics and the structure of our universe.

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